package org.usmile.algorithms.leetcode.middle;

/**
 * 79. 单词搜索
 *
 * 给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
 * 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
 *
 * 示例 1：
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
 * 输出：true
 *
 * 示例 2：
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
 * 输出：true
 *
 * 示例 3：
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
 * 输出：false
 *
 * 提示：
 * m == board.length
 * n = board[i].length
 * 1 <= m, n <= 6
 * 1 <= word.length <= 15
 * board 和 word 仅由大小写英文字母组成
 *
 * 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
 */
public class _0079 {
}

class _0079_Solution {
    private final int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    private int rows;
    private int cols;
    private boolean[][] visited;
    private char[][] board;
    private String word;

    public boolean exist(char[][] board, String word) {
        this.board = board;
        this.word = word;
        this.rows = board.length;
        this.cols = board[0].length;
        this.visited = new boolean[rows][cols];
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                if (board[row][col] == word.charAt(0)) {
                    if (dfs(row, col, 0)) {
                        return true;
                    }
                }
            }
        }

        return false;
    }

    private boolean dfs(int row, int col, int index) {
        if (board[row][col] != word.charAt(index)) {
            return false;
        }
        if (index == word.length() - 1) {
            return true;
        }

        visited[row][col] = true;
        for (int[] direction : directions) {
            int nextRow = row + direction[0];
            int nextCol = col + direction[1];
            if (nextRow >= 0 && nextRow < rows && nextCol >= 0 && nextCol < cols) {
                if (!visited[nextRow][nextCol]) {
                    if (dfs(nextRow, nextCol, index + 1)) {
                        return true;
                    }
                }
            }
        }
        visited[row][col] = false;

        return false;
    }
}